It is easy to show that if $f\colon[0,1]\to\mathbb R$ and $|f|\leq A$ and $|f''|\leq B$ then~$|f'|\leq 4A+B$. Indeed, by Taylor formula with remainder $f(x)=f(c)+(x-c)f'(c)+\frac12(x-c)^2f''(d)$ where $d$ is between $x$ and $c$. Therefore $f'(c)=\frac{f(x)-f(c)-\frac12(x-c)^2f''(d)}{x-c}$. Now let $c$ be a maximum of $|f'|$. We can assume without loss of generality that $c\geq\frac12$ (otherwise apply the same argument to the function $f(\frac12-x)$ to get the same bound). Then \begin{aligned} |f'(c)|&\leq2\left|f(0)-f(c)-\frac12(0-c)^2f''(d)\right| \\&\leq2|f(0)|+2|f(c)|+c^2 |f''(d)| \\&\leq4A+B \end{aligned} as required. In particular, $f$ is $(4A+B)$-Lipschitz.

The question is, can one obtain similar bounds for the Lipschitz constant of $f$ if one does not require differentiability but only assumes a bound on second differences: $f(x)-2f(x+h)+f(x+2h)\leq Bh^2$ ?

(asked unsuccessfully at MSE.)

It is easy to show that if $f\colon[0,1]\to\mathbb R$ and $|f|\leq A$ and $|f''|\leq B$ then~$|f'|\leq 4A+B$. Indeed, by Taylor formula with remainder $f(x)=f(c)+(x-c)f'(c)+\frac12(x-c)^2f''(d)$ where $d$ is between $x$ and $c$. Therefore $f'(c)=\frac{f(x)-f(c)-\frac12(x-c)^2f''(d)}{x-c}$. Now let $c$ be a maximum of $|f'|$. We can assume without loss of generality that $c\geq\frac12$ (otherwise apply the same argument to the function $f(\frac12-x)$ to get the same bound). Then \begin{aligned} |f'(c)|&\leq2\left|f(0)-f(c)-\frac12(0-c)^2f''(d)\right| \\&\leq2|f(0)|+2|f(c)|+c^2 |f''(d)| \\&\leq4A+B \end{aligned} as required. In particular, $f$ is $(4A+B)$-Lipschitz.

The question is, can one obtain similar bounds for the Lipschitz constant of $f$ if one does not require differentiability but only assumes a bound on second differences: $|f(x)-2f(x+h)+f(x+2h)|\leq Bh^2$ ?

(asked unsuccessfully at MSE.)